AP Physics 2 Quiz: The First Law Of Thermodynamics

20 questions · exam conditions

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The First Law Of ThermodynamicsQuestion 1 of 20

A rigid, sealed tank of gas is the system. Take $Q>0$ into the gas and $W>0$ done by the gas. The tank is heated so that $Q=+1200\,\text{J}$ is transferred to the gas; because the tank is rigid, the boundary does not move. Which statement correctly describes the gas's internal energy change?

\Delta U=0\,\text{J}

because the volume is constant so heat cannot change internal energy

\Delta U=+1200\,\text{J}

because

W=0

and

\Delta U=Q-W

\Delta U=-1200\,\text{J}

because no work is done so energy must leave as heat

\Delta U=+1200\,\text{J}

because temperature always increases by the same amount as heat added

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AP Physics 2 Quiz

AP Physics 2 Quiz: The First Law Of Thermodynamics

Practice The First Law Of Thermodynamics in AP Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on The First Law Of Thermodynamics, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\Delta U=0\,\text{J}$ because the volume is constant so heat cannot change internal energy
$\Delta U=+1200\,\text{J}$ because $W=0$ and $\Delta U=Q-W$ (correct answer)
$\Delta U=-1200\,\text{J}$ because no work is done so energy must leave as heat
$\Delta U=+1200\,\text{J}$ because temperature always increases by the same amount as heat added

Explanation: This problem tests understanding of the first law of thermodynamics. The first law states that the change in internal energy equals heat added to the system minus work done by the system: ΔU = Q - W. For a rigid tank, the boundary cannot move, so no work can be done by or on the gas: W = 0. With Q = +1200 J and W = 0, we get ΔU = (+1200 J) - 0 = +1200 J. Choice A incorrectly claims that constant volume prevents internal energy change—this is the "rigid container" misconception that confuses zero work with zero energy change. Always recognize that in rigid containers, all heat transfer goes directly to internal energy change.

Question 2

A gas in a piston-cylinder is the system. Take $Q>0$ into the gas and $W>0$ done by the gas. During a process the gas does $W=+300\,\text{J}$ of work, and its internal energy does not change ( $\Delta U=0$ ). Which statement correctly describes the heat transfer?

$Q=0\,\text{J}$ because no internal energy change implies no heat transfer
$Q=-300\,\text{J}$ because $Q=\Delta U-W$
$Q=+300\,\text{J}$ because $Q=W$ when $\Delta U=0$ (correct answer)
$Q=+300\,\text{J}$ because work done by the gas always increases its temperature

Explanation: This problem tests understanding of the first law of thermodynamics. The first law states that the change in internal energy equals heat added to the system minus work done by the system: ΔU = Q - W. Given ΔU = 0 and W = +300 J, we can solve for Q: 0 = Q - (+300 J), which gives Q = +300 J. Choice A incorrectly assumes no internal energy change means no heat transfer—this is the "zero ΔU means isolated system" misconception that ignores the possibility of balanced heat and work. Always recognize that ΔU = 0 means heat added equals work done by the system.

Question 3

A sealed piston-cylinder contains a gas (the system). Use $\Delta U=Q-W$ with $Q>0$ into the gas and $W>0$ done by the gas. In one process, $\Delta U=-120\ \text{J}$ while $Q=+50\ \text{J}$ . Which statement correctly describes the work $W$ ?

$W=+170\ \text{J}$ because more energy left as work than entered as heat (correct answer)
$W=-170\ \text{J}$ because the gas must have been compressed
$W=+70\ \text{J}$ because work equals heat minus internal energy change
$W=-70\ \text{J}$ because positive heat implies negative work

Explanation: This problem requires applying the first law of thermodynamics. The first law ΔU = Q - W can be rearranged to find work: W = Q - ΔU. Given Q = +50 J (heat added) and ΔU = -120 J (internal energy decreases), we calculate W = 50 J - (-120 J) = 50 J + 120 J = +170 J. The gas does 170 J of work, which explains why internal energy decreases despite heat input—more energy leaves as work than enters as heat. Choice C incorrectly subtracts the internal energy change instead of adding it, showing confusion about sign conventions in the rearranged equation. Always check that your answer makes physical sense: here, work output exceeds heat input, causing the internal energy decrease.

Question 4

Liquid water in a rigid, sealed container is the system. Use $\Delta U=Q-W$ with $Q>0$ into the water and $W>0$ done by the water. A heater transfers $Q=+800\ \text{J}$ to the water while the container's volume stays constant. Which statement correctly describes the energy change of the system?

$\Delta U=0$ because constant volume implies no heat transfer
$\Delta U=-800\ \text{J}$ because heat added is used entirely for expansion work
$\Delta U=+800\ \text{J}$ because $W=0$ in a rigid container (correct answer)
$\Delta U$ cannot be determined because temperature change is not given

Explanation: This problem requires applying the first law of thermodynamics. The first law states ΔU = Q - W, where Q is heat added to the system and W is work done by the system. In a rigid container, the volume cannot change, so no expansion or compression work can occur: W = 0. With Q = +800 J added to the water and W = 0, we get ΔU = 800 J - 0 = +800 J. All the heat energy goes into increasing the internal energy of the water. Choice A incorrectly assumes constant volume prevents heat transfer, confusing the constraint on work with a constraint on heat. Always identify whether the system can do boundary work before applying the first law.

Question 5

A sample of gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ with $Q>0$ into the gas and $W>0$ done by the gas. In a process, the surroundings do $350\ \text{J}$ of work on the gas, and $Q=-100\ \text{J}$ leaves the gas. Which statement correctly describes $\Delta U$ ?

$\Delta U=-450\ \text{J}$ because both heat loss and compression reduce $U$
$\Delta U=+250\ \text{J}$ because compression adds more energy than the heat loss removes (correct answer)
$\Delta U=-100\ \text{J}$ because internal energy change equals heat transfer
$\Delta U=+100\ \text{J}$ because negative heat means temperature must increase

Explanation: This problem requires applying the first law of thermodynamics. The first law relates internal energy change to heat and work: ΔU = Q - W. Here, 350 J of work is done ON the gas, which means W = -350 J (negative because the gas does negative work when compressed). With Q = -100 J (heat leaves the gas), we get ΔU = (-100 J) - (-350 J) = -100 J + 350 J = +250 J. The internal energy increases because the work done on the gas adds more energy than the heat loss removes. Choice A incorrectly treats both terms as reducing internal energy, showing the misconception that compression always decreases temperature. Always use proper sign conventions: work done BY the gas is positive, work done ON the gas is negative.

Question 6

Water in a well-insulated container is the system. Use $Q>0$ into the system and $W>0$ done by the system. A paddle wheel driven by an external motor does $300\,\text{J}$ of work on the water (so work done by the system is negative). No heat is exchanged. Which statement correctly describes $\Delta U$ for the water?

$\Delta U=0\,\text{J}$ because insulation prevents any internal energy change
$\Delta U=-300\,\text{J}$ because work done on the system reduces its internal energy
$\Delta U=+300\,\text{J}$ because $Q=0$ and $W=-300\,\text{J}$ so $\Delta U=Q-W$ (correct answer)
$\Delta U=+300\,\text{J}$ because any work implies heat is added

Explanation: This problem tests understanding of the first law of thermodynamics. The first law states that the change in internal energy equals heat added to the system minus work done by the system: ΔU = Q - W. The insulated container means Q = 0, and the paddle does 300 J of work ON the water, so W = -300 J (negative because work is done on the system). Substituting: ΔU = 0 - (-300 J) = +300 J. Choice A incorrectly assumes insulation prevents any internal energy change—this is the "insulation blocks all energy" misconception that ignores mechanical work. Always consider both heat AND work as energy transfer mechanisms.

Question 7

A gas in a sealed piston-cylinder is the system. Use $\Delta U=Q-W$ with $Q>0$ into the gas and $W>0$ done by the gas. During one process, $Q=+500\ \text{J}$ and the gas does $W=+200\ \text{J}$ on the piston. Which statement correctly describes the change in the gas's internal energy?

$\Delta U=+300\ \text{J}$ because net energy remains in the system (correct answer)
$\Delta U=+700\ \text{J}$ because heat added and work done both increase $U$
$\Delta U=-300\ \text{J}$ because doing work always decreases temperature
$\Delta U=+500\ \text{J}$ because internal energy change equals heat transfer

Explanation: This problem requires applying the first law of thermodynamics. The first law states that the change in internal energy equals heat added to the system minus work done by the system: ΔU = Q - W. Given Q = +500 J (heat added to the gas) and W = +200 J (work done by the gas), we calculate ΔU = 500 J - 200 J = +300 J. The internal energy increases by 300 J because more energy enters as heat than leaves as work. Choice B incorrectly adds Q and W instead of subtracting, suggesting the misconception that all energy transfers increase internal energy. Always remember that work done BY the system reduces its internal energy, while work done ON the system increases it.

Question 8

A fixed amount of gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ ( $Q>0$ into gas, $W>0$ by gas). During a process, the gas releases $75\ \text{J}$ of heat ( $Q=-75\ \text{J}$ ) and does $25\ \text{J}$ of work ( $W=+25\ \text{J}$ ). Which statement correctly describes $\Delta U$ ?

The gas's internal energy decreases by $100\ \text{J}$ . (correct answer)
The gas's internal energy is unchanged because heat transfer sets temperature only.
The gas's internal energy decreases by $50\ \text{J}$ .
The gas's internal energy increases by $50\ \text{J}$ because work is done by the gas.

Explanation: This problem involves the first law of thermodynamics. The first law states ΔU = Q - W, relating internal energy to heat and work. Given Q = -75 J (heat leaves the gas) and W = +25 J (work done by gas), we calculate ΔU = -75 - 25 = -100 J. The internal energy decreases by 100 J. Choice C incorrectly calculates only -50 J, perhaps by subtracting the magnitudes incorrectly or confusing the sign of work. When both heat leaves the system and work is done by the system, both terms reduce internal energy—always add their effects with proper signs.

Question 9

A gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ ( $Q>0$ into gas, $W>0$ by gas). The gas is insulated so $Q=0$ , and it expands doing $W=+150\ \text{J}$ . Which statement correctly describes the internal energy change?

The gas's internal energy increases by $150\ \text{J}$ because it expands.
The gas's internal energy decreases by $150\ \text{J}$ only if temperature decreases.
The gas's internal energy is unchanged because no heat is transferred.
The gas's internal energy decreases by $150\ \text{J}$ . (correct answer)

Explanation: This problem applies the first law of thermodynamics. The first law states ΔU = Q - W, where Q is heat added and W is work done by the system. For an insulated system (adiabatic process), Q = 0. With Q = 0 and W = +150 J (work done by expanding gas), we get ΔU = 0 - 150 = -150 J. The internal energy decreases by 150 J. Choice C incorrectly assumes that no heat transfer means no change in internal energy, ignoring that work also affects internal energy. In adiabatic processes, always remember that any work done comes entirely from the system's internal energy.

Question 10

A gas in a sealed cylinder with a movable piston is the system. Use $\Delta U=Q-W$ ( $Q>0$ into gas, $W>0$ by gas). During a process, the gas does $W=+80\ \text{J}$ , and its internal energy increases by $\Delta U=+20\ \text{J}$ . Which statement correctly describes the heat transfer $Q$ ?

Heat enters the gas: $Q=+100\ \text{J}$ . (correct answer)
Heat enters the gas: $Q=+60\ \text{J}$ .
Heat leaves the gas: $Q=-60\ \text{J}$ .
No heat is transferred: $Q=0\ \text{J}$ .

Explanation: This problem tests the first law of thermodynamics. The first law states ΔU = Q - W, which rearranges to Q = ΔU + W for finding heat transfer. Given W = +80 J (work done by gas) and ΔU = +20 J (internal energy increases), we calculate Q = 20 + 80 = +100 J. Heat enters the gas. Choice C gives only 60 J, perhaps by subtracting W from ΔU instead of adding, a sign error when rearranging the equation. Always solve for the unknown by properly rearranging the first law equation and maintaining sign consistency.

Question 11

A fixed amount of gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ , with $Q>0$ into the system and $W>0$ done by the system. During a process, $Q=+90\ \text{J}$ and $\Delta U=-30\ \text{J}$ . Which statement correctly describes the work $W$ done by the gas?

The gas does $+60\ \text{J}$ of work.
The gas does $+120\ \text{J}$ of work. (correct answer)
Work is zero because internal energy decreases.
An external agent does $+60\ \text{J}$ of work on the gas.

Explanation: This problem involves the first law of thermodynamics. The first law states ΔU = Q - W, which can be rearranged to find work: W = Q - ΔU. Given Q = +90 J (heat added) and ΔU = -30 J (internal energy decreases), we calculate W = 90 - (-30) = 90 + 30 = +120 J. The gas does 120 J of work. Choice A incorrectly subtracts instead of accounting for the negative ΔU, missing that a decrease in internal energy contributes to the work output. When solving for W, always rearrange the first law equation carefully and watch the signs.

Question 12

A gas in a cylinder with a movable piston is the system. Use $\Delta U=Q-W$ ( $Q>0$ into gas, $W>0$ by gas). During a compression, an external agent does $300\ \text{J}$ of work on the gas, so $W=-300\ \text{J}$ . At the same time, $Q=-50\ \text{J}$ (heat leaves the gas). Which statement correctly describes $\Delta U$ ?

The gas's internal energy increases by $350\ \text{J}$ because work is done on it.
The gas's internal energy increases by $250\ \text{J}$ . (correct answer)
The gas's internal energy is unchanged because heat leaves during compression.
The gas's internal energy decreases by $350\ \text{J}$ .

Explanation: This problem involves the first law of thermodynamics. The first law states ΔU = Q - W, where positive Q means heat enters the system and positive W means work is done by the system. When work is done ON the gas during compression, W = -300 J (negative because the gas does negative work). With Q = -50 J (heat leaves) and W = -300 J, we calculate ΔU = -50 - (-300) = -50 + 300 = +250 J. The internal energy increases by 250 J. Choice A incorrectly adds the magnitudes instead of using proper signs, a common error when dealing with work done on the system. Always use consistent sign conventions: work BY the system is positive, work ON the system is negative.

Question 13

A sample of gas in a piston is the system. Take $Q>0$ into the gas and $W>0$ done by the gas. The internal energy increases by $\Delta U=+400\ \text{J}$ while the gas does $W=+150\ \text{J}$ . Which statement correctly describes the heat transfer $Q$ ?

$Q=-550\ \text{J}$ because positive work requires heat to leave
$Q=+250\ \text{J}$ because some heat becomes work
$Q=+550\ \text{J}$ because heat supplies both $\Delta U$ and $W$ (correct answer)
$Q=+400\ \text{J}$ because heat equals the internal energy change

Explanation: This question tests understanding of the first law of thermodynamics. The first law can be rearranged to find heat transfer: Q = ΔU + W. Given that ΔU = +400 J (internal energy increases) and W = +150 J (work done by the gas), we calculate Q = 400 + 150 = +550 J. This positive value means heat must enter the gas to supply both the increase in internal energy and the work output. Choice A incorrectly suggests Q = +250 J, possibly by subtracting W from ΔU instead of adding, misunderstanding that both energy changes require heat input. When solving for any variable in the first law, rearrange the equation algebraically before substituting numbers.

Question 14

A gas (system = gas) is cooled while being compressed. Sign convention: $Q>0$ into system, $W>0$ done by system. During the process, $Q=-75\,\text{J}$ and $\Delta U=-25\,\text{J}$ . Which statement correctly describes the work $W$ ?

$W=+50\,\text{J}$ because $W=Q-\Delta U$
$W=+100\,\text{J}$ because the gas must do positive work whenever it cools
$W=-50\,\text{J}$ because $W=Q-\Delta U$ (correct answer)
$W=0\,\text{J}$ because compression and cooling cancel in the first law

Explanation: This problem involves the first law of thermodynamics. Given Q = -75 J (cooling) and ΔU = -25 J, we solve for W using ΔU = Q - W. Rearranging: W = Q - ΔU = -75 J - (-25 J) = -75 J + 25 J = -50 J. The negative work confirms compression (work done ON the gas). Choice A has the wrong sign. Choice C incorrectly assumes cooling requires positive work. Choice D claims impossible cancellation. Always check consistency: cooling and compression both tend to decrease volume, so they can occur together.

Question 15

A sealed cylinder contains an ideal gas (system = gas). Sign convention: $Q>0$ into system, $W>0$ done by system. During one process, $Q=+250\,\text{J}$ and the gas does $W=+150\,\text{J}$ on the piston. Which statement correctly describes the internal energy change of the gas?

$\Delta U=-100\,\text{J}$ because work done by the gas decreases temperature by $100\,\text{J}$
$\Delta U=+400\,\text{J}$ because heat added always increases internal energy by the same amount
$\Delta U=+100\,\text{J}$ because $\Delta U=Q-W$ (correct answer)
$\Delta U=+250\,\text{J}$ because internal energy change equals heat transfer

Explanation: This problem tests understanding of the first law of thermodynamics. The first law states that the change in internal energy equals heat added to the system minus work done by the system: ΔU = Q - W. Given Q = +250 J (heat into the gas) and W = +150 J (work done by the gas), we calculate ΔU = 250 J - 150 J = +100 J. Choice A incorrectly assumes heat added always equals the internal energy change, ignoring work. Choice C makes a sign error and confuses work with temperature change. Choice D ignores the work term entirely. Always apply the complete first law equation ΔU = Q - W with careful attention to sign conventions.

Question 16

A sealed piston-cylinder contains a gas (system = gas). Sign convention: $Q>0$ into system, $W>0$ done by system. In one step, $600\,\text{J}$ of heat enters the gas, and the internal energy increases by $450\,\text{J}$ . Which statement correctly describes the work done by the gas?

$W=+450\,\text{J}$ because work done equals the internal energy change
$W=+150\,\text{J}$ because $W=Q-\Delta U$ (correct answer)
$W=+1050\,\text{J}$ because work equals heat plus internal energy change
$W=-150\,\text{J}$ because increasing internal energy means work is done on the gas

Explanation: This problem tests the first law of thermodynamics. Given Q = +600 J (heat enters) and ΔU = +450 J, we find W using ΔU = Q - W. Rearranging: W = Q - ΔU = 600 J - 450 J = +150 J. The positive work means the gas expands and does work on its surroundings. Choice B incorrectly assumes increasing internal energy requires work on the gas. Choice C adds instead of subtracting. Choice D ignores the heat term. Always verify energy conservation: 600 J enters as heat, 450 J increases internal energy, and 150 J leaves as work.

Question 17

A rigid, insulated container holds a gas (system = gas). Sign convention: $Q>0$ into system, $W>0$ done by system. A paddle wheel does $300\,\text{J}$ of work on the gas. Which statement correctly describes the internal energy change of the gas?

$\Delta U=0$ because insulation prevents any change in internal energy
$\Delta U=-300\,\text{J}$ because work done on the gas must be negative internal energy
$\Delta U=+300\,\text{J}$ because $Q=0$ and $W=-300\,\text{J}$ (correct answer)
$\Delta U=+300\,\text{J}$ because temperature must increase by $300\,\text{K}$

Explanation: This problem involves the first law of thermodynamics. For an insulated container, Q = 0 (no heat transfer). The paddle wheel does 300 J of work ON the gas, which means W = -300 J in our sign convention (positive W is work done BY the system). Applying ΔU = Q - W gives ΔU = 0 - (-300 J) = +300 J. Choice A incorrectly assumes insulation prevents any energy change, but work can still change internal energy. Choice B has the wrong sign. Choice D incorrectly relates energy to temperature in Kelvin. Always identify whether work is done on or by the system to determine the correct sign.

Question 18

A gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ with $Q>0$ into the gas and $W>0$ done by the gas. The gas is heated with $Q=+200\ \text{J}$ while an external agent compresses it so the work done by the gas is $W=-50\ \text{J}$ . Which statement correctly gives $\Delta U$ ?

The gas's internal energy increases by $150\ \text{J}$ .
The gas's internal energy increases by $250\ \text{J}$ . (correct answer)
The gas's internal energy decreases by $250\ \text{J}$ .
The gas's internal energy is unchanged because heating offsets compression.

Explanation: This problem applies the first law of thermodynamics. The first law states ΔU = Q - W, where Q is heat added and W is work done by the system. Given Q = +200 J (heat added) and W = -50 J (negative because work is done ON the gas during compression), we calculate ΔU = 200 - (-50) = 200 + 50 = +250 J. The internal energy increases by 250 J. Choice A incorrectly subtracts the work magnitude, not recognizing that work done on the gas (negative W) adds to internal energy. Always remember that compression (negative W) and heating (positive Q) both increase internal energy.

Question 19

An ideal-gas sample in a piston-cylinder is the system. Use $\Delta U=Q-W$ with $Q>0$ added to the gas and $W>0$ done by the gas. The gas undergoes a slow process in which $\Delta U=0$ while it does $W=+400\ \text{J}$ . Which statement correctly describes the heat transfer $Q$ ?

Heat must leave the gas: $Q=-400\ \text{J}$ .
No heat is transferred: $Q=0$ .
Heat must enter the gas: $Q=+400\ \text{J}$ . (correct answer)
Heat must enter the gas: $Q=+800\ \text{J}$ .

Explanation: This problem tests the first law of thermodynamics. The first law states ΔU = Q - W, relating internal energy change to heat and work. Given ΔU = 0 (no change in internal energy) and W = +400 J (work done by the gas), we can solve for Q: 0 = Q - 400, so Q = +400 J. This means 400 J of heat must enter the gas to maintain constant internal energy while it does work. Choice D incorrectly doubles the heat value, perhaps confusing the relationship between Q and W. For processes with ΔU = 0 (like isothermal processes for ideal gases), always remember that Q = W.

Question 20

A gas in a piston-cylinder is the system. Use $\Delta U=Q-W$ , with $Q>0$ into the gas and $W>0$ done by the gas. During a process, $\Delta U=+40\ \text{J}$ and $Q=-10\ \text{J}$ . Which statement correctly describes the work $W$ done by the gas?

The gas does $-50\ \text{J}$ of work (work is done on the gas). (correct answer)
Work is zero because heat leaves the gas.
The gas does $+50\ \text{J}$ of work.
The gas does $+30\ \text{J}$ of work.

Explanation: This problem tests the first law of thermodynamics. The first law states ΔU = Q - W, which rearranges to W = Q - ΔU when solving for work. Given ΔU = +40 J (internal energy increases) and Q = -10 J (heat leaves), we calculate W = -10 - 40 = -50 J. The negative work means work is done ON the gas (compression). Choice A incorrectly gives positive work, perhaps misunderstanding that negative Q and positive ΔU require work input. Always check your answer's physical sense: if heat leaves but internal energy increases, work must be done on the system.